Under the action of a given coulombic force, the acceleration of an electron is 2.5×10²²m/sec². Then, the magnitude of the acceleration of a proton under the action of same force is nearly

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$1.6 \times 10^{- 19} m / \sec^{2}$

$1.5 \times 10^{19} m / \sec^{2}$

$1.6 \times 10^{27} m / \sec^{2}$

$9.1 \times 10^{31} m / \sec^{2}$

answer is C.

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Detailed Solution

$\begin{aligned} \frac{a_{1}}{a_{2}} = \frac{m_{2}}{m_{1}} \Rightarrow \frac{2.5 \times 10^{22}}{a_{2}} = 1836 \\ \Rightarrow a_{2} = 1.5 \times 10^{19} m / \sec^{2} \end{aligned}$

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